For an aqueous solution of $CaCl_2$ electrolyte,the graph between molar conductivity $(\lambda_m)$ and $(\text{concentration})^{1/2}$ is:

The molar conductivity of $0.007 \ M$ acetic acid is $20 \ S \ cm^{2} \ mol^{-1}$. What is the dissociation constant of acetic acid? (In $\times 10^{-5} \ mol \ L^{-1}$)
$[\Lambda_{H^{+}}^{\circ}=350 \ S \ cm^{2} \ mol^{-1}, \Lambda_{CH_{3}COO^{-}}^{\circ}=50 \ S \ cm^{2} \ mol^{-1}]$

Given $l/a = 0.5 \ cm^{-1}$,$R = 50 \ \Omega$,$N = 1.0$. The equivalent conductance of the electrolytic cell is ................ $\Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.

The conductivity of an electrolytic solution decreases on dilution due to

The resistance of a $1 \, N$ solution of a salt is $50 \, \Omega$. The two platinum electrodes in the solution are $2.1 \, cm$ apart and each has an area of $4.2 \, cm^2$. Calculate the equivalent conductivity of the solution.

Resistance of a conductivity cell filled with $0.1 \ mol \ L^{-1}$ $NaCl$ is $100 \ \Omega$. If the resistance of the same cell when filled with $0.02 \ mol \ L^{-1}$ $NaCl$ solution is $258 \ \Omega$,the conductivity of $0.02 \ mol \ L^{-1}$ $NaCl$ solution is (Conductivity of $0.1 \ mol \ L^{-1}$ $NaCl$ is $1.29 \ S \ m^{-1}$) (in $S \ m^{-1}$)